Assume that the production of items follows a Berou scheme of trials, and the pr

Question

Assume that the production of items follows a Berou scheme of trials, and the probability that an iten is defective is p= 0.015. Consider boxes that con- tain n = 100 items. (a) Using Poisson approximation, find approximate value for the probability of finding k defective items. Compare to exact probabilities for snall values of k say k 0, k 1, and k=2. (b) Assume that we are not sure about the value of p, but it could be any value in the following set: 0.005,0.015,0.025) We open a box and find 2 defective items. What should we believe the value of p is? Explain your answer.

Explanation / Answer

(a) Using poisson approiximation,

Average number of defective piece in n= 100 items () = 100 * 0.015 = 1.5

Pr( X = k ; ) = e- k / k!

so,

Pr( X = 0 ; 1.5) = e-1.5 (1.5)0 / 0! = 0.2231

Pr( X = 1 ; 1.5) = e-1.5 (1.5)1 / 1! = 0.3347

Pr( X = 2 ; 1.5) = e-1.5 (1.5)2 / 2! = 0.2510

(b) P can be {0.005, 0.015, 0.025}

We find two defective items out of 100.

so Pr(X = 2; 100; 0.005) = POISSON ( X = 2; 0.5) = e-0.5 (0.5)2 / 2! = 0.0758

Pr(X = 2; 100; 0.015) = POISSON ( X = 2; 1.5) = e-1.5 (1.5)2 / 2! = 0.2510

Pr(X = 2; 100; 0.025) = POISSON ( X = 2; 2.5) = e-2.5 (2.5)2 / 2! = 0.2565

so here we found that for p = 0.015 and 0.025 the probability of getting two defective items out of 100, is apprxomiatley same with some edge to p = 0.025.

SO we will believe that the probaiblity of defective item in the box is 0.025

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